Nontrivial cycles in the zero set of a map

Question

Let $B_4$ be a 4-disc, $X:=S^1\times B_4$, $\partial X=S^1\times S^3$ and $f: X\to\mathbb{R}^3$ be continuous, such that $f|_{\partial X}: (s_1, s_2)\mapsto H(s_2)$, where $H: S^3\to S^2\subseteq\mathbb{R}^3$ is the Hopf fibration. How to show that $H_1(f^{-1}(0))\stackrel{i_*}{\longrightarrow} H_1(X)$ is surjective (and is it necessarily true)?

(I appologize that it looks so technical but it is closely connected to a deeper problem: if I know a sphere-valued map $f$ on the boundary of some space $X$, then the zero set of possible $\mathbb{R}^n$-valued extensions correspond to solutions of $1$-perturbations (wrt. some norm) of the equation $f(x)=0$. My goal is to describe the topology of the solution set(s). In this example, does it need to contain a nontrivial circle?)

Answer 1

It seems that the result is not true, at least if $H_1$ means singular homology.

The basic idea of the counterexample is to construct $f$ so that $f^{-1}(\{0\})$ doesn't contain an ordinary circle, but something like a Warsaw circle instead. This leaves open the possibility that the result might still be true for some other homology theory (Čech homology? Strong homology?) or if we restrict to a nicer class of functions (Morse functions?)

Let $X=A\cup B$, where $$A=\left\{(x,y)\in S^1\times D^4\mid \|y\|\geq\frac12\right\}$$ and $$B=\left\{(x,y)\in S^1\times D^4\mid\|y\|\leq\frac12\right\}.$$

Now, for $(x,y)\in A$, define $$f(x,y)=\left(\frac{3+x_1}2\|y\|-\frac{1+x_1}2\right)H\left(\frac{y}{\|y\|}\right),$$ where $x_1$ is the first coordinate of $x=(x_1,x_2)\in\mathbb R^2$. Note that $f$ is continuous on $A$. If $\|y\|=1,$ we have $f(x,y)=H(y)$, as required. Furthermore, if $(x,y)\in A\cap B$, i.e. if $\|y\|=\frac12$, we have $$f(x,y)=\frac{1-x_1}4H\left(\frac y{\|y\|}\right).$$

We proceed to define $f(x,y)$ for $(x,y)\in B$. First, define $q:(0,\infty)\to\mathbb R^4$ by $$q(t)=\left(\frac{\sin\frac1{t}}{2|\sin\frac1t|-8},0,0,0\right)$$ and $\phi:(0,\infty)\times\mathbb R^4\to\mathbb R^4$ by $$\phi(t,y)=y+q(t)-2\|y\|q(t).$$

The function $\phi$ has the nice properties that $\|\phi(t,y)\|\leq\frac12$ if $\|y\|\leq\frac12$ and $\phi(t,y)=y$ for $\|y\|=\frac12$. Also, for each fixed $t$, $y\mapsto\phi(t,y)$ is bijective. (Edit: Details provided at the end.) Furthermore, $$\phi\left(t,\left(\frac18\sin\frac1t,0,0,0\right)\right)=0.$$

Now, define $$f(x,y)=\frac{1-x_1}2 \|\phi(1-x_1,y)\|H\left(\frac{\phi(1-x_1,y)}{\|\phi(1-x_1,y)\|}\right).$$ Note that this is undefined for $x_1=1$ but it can be continuously extended by defining $$f((1,0),y)=0.$$ This defines $f(x,y)$ for $(x,y)\in B$. For this to make sense, we have to verify that the two definitions agree for $(x,y)\in A\cap B$. So, let $\|y\|=\frac12$. Then, $$f(x,y)=\frac{1-x_1}2\|y\|H\left(\frac y{\|y\|}\right),$$ by our observation that $\phi(t,y)=y$ for $\|y\|=\frac12$. This means that the two definitions indeed agree on $A\cap B$, so $f$ is well-defined.

Finally, we compute $f^{-1}(\{0\})$. First, for $(x,y)\in A$, $f(x,y)=0$ is possible only if $$\frac{3+x_1}2\|y\|-\frac{1+x_1}2=0,$$ which means that $$\|y\|=\frac{1+x_1}{3+x_1}=1-\frac2{3+x_1}\leq 1-\frac12=\frac12.$$ So, it is sufficient to find all $(x,y)\in B$ such that $f(x,y)=0$.

We have already observed that $\{(1,0)\}\times D^4\subseteq f^{-1}(\{0\})$, so assume $x_1\neq 1$. Then, $f(x,y)=0$ implies that $$\|\phi(1-x_1,y)\|=0$$ or eqivalently, that $$\phi(1-x_1,y)=0.$$ By our observations about $\phi$, this holds precisely if $$y=\left(\frac18\sin\frac1{1-x_1},0,0,0\right).$$ In other words, $f^{-1}(\{0\})$ is a Warsaw circle, with the usual interval at the end replaced by a $4$-disk. Therefore, $H_1(f^{-1}(\{0\})$ is trivial and we are done.

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Added: Here are some more details about the function $\phi$, to make things easier to read. First, observe that $$\|q(t)\|=\frac{|\sin\frac1t|}{8-2|\sin\frac1t|}=-\frac12+\frac4{8-2\sin\frac1t}\leq -\frac12+\frac46=\frac16.$$ We claim that for any vector $q\in\mathbb R^4$ such that $\|q\|<\frac12$, the function $\psi:\mathbb R^4\to\mathbb R^4$, defined by $$\psi(y)=y+q-2\|y\|q$$ is bijective, takes the $4$-disk $\{y\mid\|y\|\leq\frac12\}$ into itself and is equal to the identity when restricted to the $3$-sphere $\{y\mid\|y\|=\frac12\}$. The last of these is obvious.

To see that the second last is true, take $\|y\|\leq\frac12$ and compute: $$\big\|y+(1-2\|y\|)q\big\|\leq\|y\|+(1-2\|y\|)\|q\|\leq\|y\|+(1-2\|y\|)\frac12=\frac12.$$

It remains to show that $\psi$ is bijective. Surjectivity is left to the reader, since we don't really use it anywhere. Injectivity is verified by a simple calculation: $$y_1+q-2\|y_1\|q=y_2+q-2\|y_2\|q$$ is equivalent to $$y_1-y_2=2(\|y_1\|-\|y_2\|)q.$$ If the two norms are equal, we have $y_1=y_2$ and we are done, otherwise the equation is equivalent to $$\frac{y_1-y_2}{\|y_1\|-\|y_2\|}=2q.$$ After taking norms, this becomes $$2\|q\|=\frac{\|y_1-y_2\|}{\big|\|y_1\|-\|y_2\|\big|}\geq 1$$ by the triangle inequality. This contradicts the fact that $\|q\|<\frac12$.

One last observation: $$\psi\left(-\frac{q}{1+2\|q\|}\right)=0.$$ This translates to the fact that $$\phi\left(t,\left(\frac18\sin\frac1t,0,0,0\right)\right)=0.$$ Thus, all properties of $\phi$ we needed are verified.

Answer 2

I think the result is not true.

First, note that since $B^4$ is perfectly normal, for some $\epsilon < 1/2$, we can find a function $g \colon B^4 \rightarrow [0,\epsilon/2]$ which is $0$ along the boundary $S^3$, $0$ at the two points $\left(\pm \frac{1}{2},0,0,0\right)$, $\epsilon/2$ on a small ball of radius $\epsilon$ around the origin, and otherwise not $0$ or $\epsilon/2$ (by an extended version of Urysohn's Lemma).

Now let $R_{\pi} \colon B^4 \rightarrow B^4$ be the rotation by $\pi$ around the first two coordinates. Then the function $h \colon B^4 \rightarrow [0,1]$ given by $\frac{1}{2}(g + g\circ R_{\pi}$ satisfies the same properties of $g$, but is also $R_{\pi}$-invariant.

Consider the function $\widetilde{f} \colon \mathbb{R} \times B^4 \rightarrow \mathbb{R}^3$ given by sending $(\theta,v=(v_1,v_2,v_3,v_4))$ to $$\frac{\|v+(1/2\cos \theta,1/2\sin\theta,0,0)\| \cdot \|v - (1/2\cos\theta,1/2\sin\theta,0,0)\|}{\sqrt{\frac{5}{4}-v_1\cos\theta-v_2\sin\theta}\cdot \sqrt{\frac{5}{4}+v_1\cos\theta+v_2\sin\theta}}H\left(\frac{f \cdot (0,0,0,1) + v}{\|f \cdot (0,0,0,1) + v\|}\right).$$ It is easy to check that this is well-defined (i.e. the denominators are never zero and the input to $H$ has norm $1$). Further, when $\|v\| = 1$, it is easy to check that this map is just the Hopf map $H$. And finally, this is invariant under $\theta \mapsto \theta + \pi$, so this descends to a map $$f \colon X \rightarrow \mathbb{R}^3$$ satisfying the necessary conditions. However, $f^{-1}(0)$ consists of points of the form $$(\theta, \pm(1/2\cos\theta,1/2\sin\theta,0,0))$$ as $\theta$ varies. In $X$, this is a circle which wraps around the $S^1$-coordinate twice, and in particular, we have $i_* \colon H_1(f^{-1}(0)) \rightarrow H_1(X)$ is a doubling map $\mathbb{Z} \rightarrow \mathbb{Z}$, and is therefore not surjective.